Pre-Algebra Examples

$p (x) = 5 x^{4} - 3 x^{3} + 6 x^{2} - 5$

Step 1

Check the leading coefficient of the function. This number is the coefficient of the expression with the largest degree.

Largest Degree: $4$

Leading Coefficient: $5$

Step 2

Simplify each term.

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Step 2.1

Cancel the common factor of $5$ .

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Step 2.1.1

Cancel the common factor.

$p (x) = \frac{5 x^{4}}{5} + \frac{- 3 x^{3}}{5} + \frac{6 x^{2}}{5} + \frac{- 5}{5}$

Step 2.1.2

Divide $x^{4}$ by $1$ .

$p (x) = x^{4} + \frac{- 3 x^{3}}{5} + \frac{6 x^{2}}{5} + \frac{- 5}{5}$

Step 2.2

Move the negative in front of the fraction.

$p (x) = x^{4} - \frac{3 x^{3}}{5} + \frac{6 x^{2}}{5} + \frac{- 5}{5}$

Step 2.3

Divide $- 5$ by $5$ .

$p (x) = x^{4} - \frac{3 x^{3}}{5} + \frac{6 x^{2}}{5} - 1$

Step 3

Create a list of the coefficients of the function except the leading coefficient of $1$ .

$- \frac{3}{5}, \frac{6}{5}, - 1$

Step 4

There will be two bound options, $b_{1}$ and $b_{2}$ , the smaller of which is the answer. To calculate the first bound option, find the absolute value of the largest coefficient from the list of coefficients. Then add $1$ .

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Step 4.1

Arrange the terms in ascending order.

$b_{1} = | - \frac{3}{5} |, | - 1 |, | \frac{6}{5} |$

Step 4.2

The maximum value is the largest value in the arranged data set.

$b_{1} = | \frac{6}{5} |$

Step 4.3

$\frac{6}{5}$ is approximately $1.2$ which is positive so remove the absolute value

$b_{1} = \frac{6}{5} + 1$

Step 4.4

Write $1$ as a fraction with a common denominator.

$b_{1} = \frac{6}{5} + \frac{5}{5}$

Step 4.5

Combine the numerators over the common denominator.

$b_{1} = \frac{6 + 5}{5}$

Step 4.6

Add $6$ and $5$ .

$b_{1} = \frac{11}{5}$

Step 5

To calculate the second bound option, sum the absolute values of the coefficients from the list of coefficients. If the sum is greater than $1$ , use that number. If not, use $1$ .

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Step 5.1

Simplify each term.

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Step 5.1.1

$- \frac{3}{5}$ is approximately $- 0.6$ which is negative so negate $- \frac{3}{5}$ and remove the absolute value

$b_{2} = \frac{3}{5} + | \frac{6}{5} | + | - 1 |$

Step 5.1.2

$\frac{6}{5}$ is approximately $1.2$ which is positive so remove the absolute value

$b_{2} = \frac{3}{5} + \frac{6}{5} + | - 1 |$

Step 5.1.3

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$b_{2} = \frac{3}{5} + \frac{6}{5} + 1$

Step 5.2

Combine fractions.

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Step 5.2.1

Combine the numerators over the common denominator.

$b_{2} = 1 + \frac{3 + 6}{5}$

Step 5.2.2

Simplify the expression.

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Step 5.2.2.1

Add $3$ and $6$ .

$b_{2} = 1 + \frac{9}{5}$

Step 5.2.2.2

Write $1$ as a fraction with a common denominator.

$b_{2} = \frac{5}{5} + \frac{9}{5}$

Step 5.2.2.3

Combine the numerators over the common denominator.

$b_{2} = \frac{5 + 9}{5}$

Step 5.2.2.4

Add $5$ and $9$ .

$b_{2} = \frac{14}{5}$

Step 5.3

Arrange the terms in ascending order.

$b_{2} = 1, \frac{14}{5}$

Step 5.4

The maximum value is the largest value in the arranged data set.

$b_{2} = \frac{14}{5}$

Step 6

Take the smaller bound option between $b_{1} = \frac{11}{5}$ and $b_{2} = \frac{14}{5}$ .

Smaller Bound: $\frac{11}{5}$

Step 7

Every real root on $p (x) = 5 x^{4} - 3 x^{3} + 6 x^{2} - 5$ lies between $- \frac{11}{5}$ and $\frac{11}{5}$ .

$- \frac{11}{5}$ and $\frac{11}{5}$